C Pointers – Fixing “Assignment Discards ‘const’ Qualifier” Error

c++constantspointers

In the following function:

char *mystrtok(const char *input, const char *delim,char *rest) {
    int i;
    for (i = 0; input[i] != *delim && input[i] != '\0'; ++i) {
        continue;
    }
    char *result = malloc(sizeof(char) * (i + 2));
    memcpy(result, input, i + 1);
    result[i + 1] = '\0';
    if (input[i + 1] != '\0') 
        rest = input + i + 2;
    else
        rest = NULL;
    return result;
}

I am getting assignment discards 'const' qualifier from pointer target type for the line rest = input + i + 2, however, as you can see, rest is not a constant pointer. What am I doing wrong here?

Best Answer

input is a pointer to a constant char, and you're assigning it to a pointer to a non-constant char. This here might be an interesting reading for you.

Related Solutions

C Pointers – Understanding ‘const’ and Double Pointers Compatibility

A conversion from e.g. char * to const char * is always safe. Through the const char *, the data that's pointed to can't be modified, and that's it.

On the other hand, a conversion from char ** to const char ** can be unsafe, therefore it isn't allowed implicitly. Instead of explaining it, consider the following code:

void foo(const char **bar)
{
    const char *str = "test string";
    *bar = str; // perfectly legal
}

int main(void)
{
    char *teststr[] = {0};
    foo((const char **)teststr);
    // now teststr points to a `const char *`!

    *teststr[0] = 'x'; // <- attempt to modify read-only memory
                       //    ok in this line, there's no const qualifier on teststr!
}

If the conversion from char ** to const char ** when calling foo() would be implicit, you would have an implicit way of converting a const away.

Best Answer

Related Solutions

C Pointers – Understanding ‘const’ and Double Pointers Compatibility

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